1. Repunit Lehmer numbers
Javier Cilleruelo
Instituto de Ciencias Matem´ticas (CSIC-UAM-UC3M-UCM)
a
and Departamento de Matem´ticas, Facultad de Ciencias
a
Universidad Aut´noma de Madrid
o
28049, Madrid, Espa˜a n
franciscojavier.cilleruelo@uam.es
Florian Luca
Instituto de Matem´ticas
a
Universidad Nacional Autonoma de M´xico
e
C.P. 58089, Morelia, Michoac´n, M´xico
a e
fluca@matmor.unam.mx
Abstract
A Lehmer number is a composite positive integer n such that φ(n) |
n − 1. In this paper, we show that given a positive integer g > 1 there
are at most finitely many Lehmer numbers which are repunits in base
g, and they are all effectively computable. Our method is effective and
we illustrate it by showing that there is no such Lehmer number when
g ∈ [2, 1000].
2000 Mathematics Subject Classification: Primary 11N25, Secondary 11B39
Keywords: Lehmer numbers, repunits
1 Introduction
Let φ(n) be the Euler function of the positive integer n. Clearly, φ(n) = n−1
if n is a prime. Lehmer [4] (see also B37 in [3]) conjectured that if φ(n) | n−1,
then n is prime. To this day, no counterexample to this conjecture has been
found. A composite number m such that φ(m) | m − 1 is called a Lehmer
number. Thus, Lehmer’s conjecture is that Lehmer numbers don’t exist but
it is not even known that there should be at most finitely many of them.
Given a positive integer g > 1 a base g repunit is a number of the form
m = (g n − 1)/(g − 1) for some integer n ≥ 1. We will refer to such numbers
1

2. simply as repunits without mentioning the dependence on g. It is not known
whether given g there are infinitely many repunit primes. When g = 2 such
primes are better known as Mersenne primes. In [5], it was shown that there
is no Lehmer number in the Fibonacci sequence. Here, we use some ideas
from [5] together with finer arguments to prove the following results. In
what follows, we write un = (g n − 1)/(g − 1).
Theorem 1. For each fixed g > 1, there are only finitely many positive inte-
gers n such that un is a Lehmer number, and all are effectively computable.
Theorem 2. There is no Lehmer number of the form un when 2 ≤ g ≤ 1000.
Acknowledgement. We thank the anonymous referee for numerous comments
which improved the quality of this paper. Work on this paper was done during a
pleasant visit of F. L. at the Mathematics Department of the Universidad Aut´noma
o
de Madrid in Spring of 2008. He thanks the people of that Institution for their
hospitality. J. C. was supported in part by Project MTM2005-04730 from MEC
(Spain) and the joint Madrid Region-UAM project TENU2 (CCG07-UAM/ESP-
1814). F. L. was also supported in part by Grants SEP-CONACyT 79685 and
PAPIIT 100508.
2 Preliminaries
For a prime q and a nonzero integer m we write νq (m) for the exponent
of q in the factorization of m. We start by collecting some elementary and
well-known properties of the sequence of general term un = (g n − 1)/(g − 1)
for n ≥ 1.
Lemma 1. i) un = g n−1 + · · · + g + 1. In particular, un is coprime to g.
ii) The sequence un satisfies the linear recurrence
u1 = 1, un = gun−1 + 1, n ≥ 2. (1)
iii) If d | n, then ud | un .
iv) Let q be a prime. If q | n, then q | φ(un ).
v) Let q be a prime not dividing g. If q | n, then νq (un−1 ) ≤ νq (uf ) ≤
νq (uq−1 ), where f is the order of g modulo q.
vi) If un is a Lehmer number, then (un , g − 1) = 1.
2

3. Proof. i) and ii) are obvious. For iii), we observe that
gn − 1 (g d )n/d − 1 g d − 1 n
un = = · = (g d ) d −1 + · · · + 1 ud .
g−1 gd − 1 g−1
iv) If q = 2, then un ≥ u2 = g + 1 > 2, therefore φ(un ) is even. Assume
now that q is odd. Let p be a prime which divides uq . Then, g q ≡ 1 (mod p),
so the order of g modulo p is 1 or q. If it is q, then q | p − 1 | φ(uq ). Since
by iii ) we know that uq | un , we get that q | φ(uq ) | φ(un ), which is what
we wanted. Assume now that the order of g modulo p is 1 for all primes p
dividing uq . Let us show that this cannot happen. If it would, then p | g − 1
for all such primes p. Since also p | uq , we have
gq − 1
0 ≡ uq ≡ = g q−1 + · · · + g + 1 ≡ 1 + · · · + 1 + 1 ≡ q,
g−1
where all congruences above are modulo p. Thus, p | q, therefore p = q.
Hence, uq = q α for some positive integer α. However, writing g − 1 = qλ
with some positive integer λ, we get
uq = (1 + qλ)q−1 + (1 + qλ)q−2 + · · · + (1 + qλ) + 1
≡ (1 + (q − 1)qλ) + (1 + (q − 2)qλ) + · · · + (1 + qλ) + 1 (mod q 2 )
≡ q + qλ((q − 1) + · · · + 1) (mod q 2 )
q 2 (q − 1)λ
≡ q+ (mod q 2 )
2
≡ q (mod q 2 ).
In the above chain of congruences, we used the fact that q is odd, therefore
(q − 1)/2 is an integer. The above argument shows that q uq ; hence, α =
1. So, uq = q. However, we clearly have uq ≥ 2q − 1 > q, which is a
contradiction.
v) We may also assume that q | un−1 , otherwise νq (un−1 ) = 0 and the
first inequality is clear. Now g n−1 ≡ 1 (mod q), and so f | n − 1. We now
write n−1
−1
un−1 = (g f ) f + · · · + 1 uf .
The quantity in brackets above is not divisible by q since it is congruent to
(n − 1)/f modulo q and q | n. Thus, νq (un−1 ) ≤ νq (uf ) ≤ νq (uq−1 ), where
the last inequality follows because f | q − 1, so, uf | uq−1 by iii).
vi) Suppose that q is a prime dividing both un and g − 1. We then have
that g ≡ 1 (mod q) and un = g n−1 + · · · + 1 ≡ n (mod q). Thus, q | n. By
3

4. iv), we know that q | φ(un ). Since un is a Lehmer number, we know that
φ(un ) | un − 1 = gun−1 . Since q divides g − 1, it cannot divide g, therefore
q | un−1 . Hence, q | un − un−1 = g n−1 , which is not possible.
In the next lemma, we gather some known facts about Lehmer numbers.
Lemma 2. i) Any Lehmer number must be odd and square-free.
K
ii) If m = p1 · · · pK is a Lehmer number, then K 2 > m.
iii) If m = p1 · · · pK is a Lehmer number, then K ≥ 14.
Proof. i) If m > 2 then φ(m) is even, and since φ(m) | m − 1, we get that
m must be odd. If p2 | m, then p | φ(m), and since φ(m) | m − 1, we have
p | m − 1, which is not possible. Part ii) was proved by Pomerance in [6],
while part iii) was proved by Cohen and Hagis in [2].
Lemma 3. Theorems 1 and 2 hold when g is even.
Proof. Note that
2K | (p1 − 1) · · · (pK − 1) = φ(un ) | un − 1 = gun−1 .
We observe that if g is even, then un−1 is odd. In that case, we have
K ≤ ν2 (φ(un )) ≤ ν2 (gun−1 ) = ν2 (g), (2)
implying, by Lemma 2 ii), that
K ν2 (g)
g n−1 < un < K 2 ≤ (ν2 (g))2 ≤ (ν2 (g))g .
Thus,
g log(ν2 (g))
n≤1+ .
log g
For Theorem 2, we observe that ν2 (g) ≤ 9 for any g ≤ 1000, and we obtain
a contradiction from (2) and Lemma 2 iii).
From Lemma 1 i), we see that if g is odd and n is even, then un is
even, so Lemma 2 i) shows that un cannot be a Lehmer number. From now
on, we shall assume that both g and n are odd and larger than 1 and that
un = (g n − 1)/(g − 1) is a Lehmer number. We also keep the notation:
α α
n = q1 1 · · · qs s , where 2 < q1 < · · · < qs (3)
are primes and α1 , . . . , αs are positive integers, and
un = p1 · · · pK , where 2 < p1 < · · · < pK (4)
are also primes.
4

5. 3 Proof of Theorem 1
3.1 Primitive divisors
Let (An )n≥1 denote a sequence with integer terms. We say that a prime p is
a primitive divisor of An if p | An and gcd(p, Am ) = 1 for all non-zero terms
Am with 1 ≤ m < n.
In 1886, Bang [1] showed that if g > 1 is any fixed integer, then the
sequence (An )n≥1 of nth term An = g n − 1 has a primitive divisor for any
index n > 6.
We will apply this important theorem to our sequence un .
Lemma 4. If d > 1 is odd, then ud has a primitive divisor pd . Furthermore,
pd ≡ 1 (mod 2d).
Proof. We revisit the argument already used at Lemma 1 iv). We write
vn = g n − 1. It is well-known that gcd(vn , vm ) = vgcd(n,m) . Observe also
that
vd
= ud = g d−1 + · · · + 1 ≡ d (mod g − 1),
v1
therefore if d is a prime not dividing g − 1, then vd has primitive divisors. If
d > 2 is a prime dividing g − 1, then the above argument, or the argument
from the proof of Lemma 1 iv ), shows that gcd(vd , v1 ) is a power of d. Write
g − 1 = dλ and observe that
vd
= (1 + dλ)d−1 + (1 + dλ)d−2 + · · · + 1
v1
≡ (1 + (d − 1)dλ) + (1 + (d − 2)dλ) + · · · + 1
= d + dλ((d − 1) + (d − 2) + · · · + 1) (mod d2 )
d2 (d − 1)
≡ d+ λ (mod d2 ) ≡ d (mod d2 ).
2
Thus, d vd /v1 , and therefore
vd 1
= (g d−1 + · · · + 1) > 1
dv1 d
is an integer coprime to v1 , so vd again has primitive divisors. Thus, v3 and
v5 (and, of course, v1 if g > 2) have primitive divisors. The fact that vd has
primitive divisors for all odd d ≥ 7 follows from Bang’s result.
We now note that if p is a primitive prime divisor of vd for d > 1, then
g d ≡ 1 (mod p), and d is the order of g (mod p). Indeed, for if not, then
5

6. f < d and p | vf , contradicting the fact that p is primitive for vd . So, d | p−1,
and since d is odd, we get that d | (p − 1)/2. Thus, p ≡ 1 (mod 2d).
Since a prime factor of g − 1 cannot be a primitive divisor for vd except
for d = 1, we deduce that if d > 1, then the primitive prime divisors for vd
are exactly those of ud = vd /(g − 1), and we get the first assertion of the
lemma.
In what follows, for a positive integer m we use ω(m) and τ (m) for the
number of prime divisors and the total number of divisors of m, respectively.
Lemma 5. If un is square-free, n is odd and (un , g − 1) = 1, then
un ω(n) q log g
log < 1 + log
φ(un ) 2q log(2q + 1)
τ (n) − 2 q 2 log g
+ 1 + log ,
2q 2 log(2q 2 + 1)
where q is the smallest prime dividing n.
Proof. We write Pd = {p is primitive prime divisor for ud }. We shall first
prove that
:= p = un .
1<d|n p∈Pd
To see the above formula, we observe that if p | ud and p g − 1, then p ∈ Pd
for some 1 < d | n. Since un is square-free, we have that un | . On the
other hand, the sets Pd are disjoint, and if p ∈ Pd , then p | ud | un . Thus,
| un .
Now, since un is square-free,
φ(un ) = (p − 1),
1<d|n p∈Pd
and then
un 1
log < .
φ(un ) p−1
d|n p∈Pd
d>1
Since all the primes p ∈ Pd are congruent to 1 (mod 2d), we have
#Pd
1 1 1 1
Sd := ≤ ≤ (1 + log #Pd ).
p−1 2d j 2d
p∈Pd j=1
6

7. To bound the cardinality of Pd , we observe that (2d + 1)#Pd ≤ ud < g d , so
d log g
#Pd < .
log(2d + 1)
We observe that d ≥ q and if d is not a prime, then d ≥ q 2 . Then
1 q log g
Sd = Sd + Sd ≤ ω(n) 1 + log
2q log(2q + 1)
1<d|n d|n d|n
d prime d composite
1 q 2 log g
+(τ (n) − 2) 1 + log .
2q 2 log(2q 2 + 1)
3.2 Bounds for q1 and τ (n)
Recall that we keep the notations from (3) and (4).
Lemma 6. If un is a Lehmer number and n is odd, then
αi (αi + 1) α
τ (n/qi ) ≤ τ (n/qi i ) ≤ νqi (φ(un )) ≤ νqi (gun−1 )
2
νqi (g), if qi |g;
≤ (5)
νqi (uqi −1 ), if qi g
for all i = 1, . . . , s.
Proof. Lemma 4 implies that for each divisor of n of the form qi d withα
αi
1 ≤ α ≤ αi and d | (n/qi ), the divisor uqi d of un has a primitive prime
α
factor pqi d ≡ 1 (mod dqi
α
α ). In particular, q α | p α − 1, and the primes p α
i dqi dqi
α
are distinct as d ranges over the divisors of n/qi i . Thus,
α
(1+···+αi )τ (n/qi i )
qi | (pdqi − 1) |
α (p − 1)
1≤α≤αi d|n/q αi p|un
i
= φ(un ) | un − 1 | gun−1 ,
which gives the two central inequalities. The first inequality is trivial and
the equality holds when αi = 1. When qi | g, the last inequality follows from
Lemma 1 i), while when qi g, then νqi (gun−1 ) = νqi (un−1 ), and we apply
Lemma 1 v) to get the desired conclusion.
7

8. √
Lemma 7. Let un be a Lehmer number with both n and g odd. If qi > g,
then
τ (n/qi ) ≤ qi − 2.
√
Proof. If qi | g and qi > g, then νqi (g) = 1, and Lemma 6 above gives
τ (n/qi ) ≤ νqi (g) = 1 ≤ qi − 2. (6)
If qi g, then, again by Lemma 6 above, we have
τ (n/qi ) ≤ νqi (uqi −1 ).
Observe that
uqi −1 | g qi −1 − 1 = g (qi −1)/2 − 1 g (qi −1)/2 + 1 .
Since qi cannot divide both factors above, we have that
τ (n/qi ) ≤ νqi (g (qi −1)/2 + ) for some ∈ {−1, +1}.
If τ (n/qi ) ≥ qi − 1, then
qi i −1 ≤ qi
q τ (n/qi )
≤ g (qi −1)/2 + 1 ≤ (qi − 1)(qi −1)/2 + 1,
2
(7)
and we get a contradiction for qi > 3, because
qi i −1 = ((qi − 1) + 1)(qi −1)/2
q 2
and the expression on the right is larger that (qi − 1)(qi −1)/2 + 1 except when
2
qi = 3.
2
Finally, if qi = 3, the only odd g < qi with qi g are g = 5 and g = 7.
But in both cases we have τ (n/3) ≤ ν3 (u2 ) ≤ 1 ≤ qi − 2, which completes
the proof of this lemma.
Lemma 8. Let un be a Lehmer number with both n and g odd. Then
√
q1 ≤ max{ g, 19}. (8)
Proof. Assume that the above inequality does not hold. Then q1 ≥ 23,
2 √
g ≤ q1 − 1, and since q1 > g, we can apply Lemma 7 to deduce that
τ (n) ≤ 2τ (n/q1 ) ≤ 2q1 − 4. We also observe that τ (n) ≥ 2ω(n) , so ω(n) ≤
log(2q1 − 4)/ log 2.
8

9. Since un is a Lehmer number, we have that 2 ≤ un /φ(un ). Now Lemma
5 and the bounds above give
log ((2q1 − 4)/ log 2) 2
q1 log(q1 − 1)
log 2 < 1 + log
2q1 log(2q1 + 1)
2q1 − 6 q 2 log(q 2 − 1)
+ 2 1 + log 1 2
1
,
2q1 log(2q1 + 1)
which is false for q1 ≥ 23.
For a given value of g, Lemma 8 gives us our bound for q1 and then
this is used in Lemma 6, since τ (n) ≤ 2τ (n/q1 ), to give a bound for τ (n).
Observe also that ω(n) ≤ log τ (n)/ log 2.
3.3 The conclusion of the proof of Theorem 1
Since we have already proved that both s = ω(n) and τ (n) are bounded by
effectively computable constants depending only on g, in order to conclude
the proof of Theorem 1 it is enough to prove that all the primes qi with
i = 1, . . . , s are also bounded by effectively computable constants depending
on g. We shall prove this by induction on i = 1, . . . , s observing that this has
already been achieved for i = 1. Let i ≤ s − 1 and assume that qi has been
j=i α
bounded. Put Qi = j=1 qj j . There are only finitely many possibilities for
this number. We put gi = g Qi , ni = n/Qi and rewrite the condition that un
is Lehmer as
n n
g Qi − 1 gi i − 1 g Qi − 1 gi i − 1
aφ · = un − 1 = · −1
g−1 gi − 1 g−1 gi − 1
m
with some integer a ≥ 2. We put wm = (gi − 1)/(gi − 1) for the sequence
of repunits in base gi . Then, since un is square-free, we get that
aφ(uQi )φ(wni ) = uQi wni − 1,
therefore
φ(uQi ) wni 1
a = − . (9)
uQi φ(wni ) uQi φ(wni )
The left hand side takes only finitely many values, which are all effectively
computable. Assume that it takes some value δ ≤ 1. Then
1
wni − 1 < wni − = δφ(wni ) ≤ φ(wni ),
uQi
9

10. a contradiction. Thus, it remains to study the case when the right hand side
in (9) is > 1. Let δi > 1 be the smallest possible value larger than 1 of the
left hand side of (9). Clearly, this is effectively computable. We then get
wni
δi < .
φ(wni )
We observe that wni is a sequence “like” un but the new value of g is
gi = g Qi and the new value of n is ni = n/Qi . Thus, the smallest prime
factor of ni is qi+1 . We also note that τ (ni ) = τ (n/Qi ) < τ (n) which is
bounded, and that ω(ni ) < ω(n). Finally, we observe that (wni , g Qi −1) = 1,
otherwise, since (wni , g−1) = 1, the number un = (g Qi −1)wni /(g−1) would
not be square-free.
We now apply Lemma 5 to obtain that
ω(ni ) Qi qi+1 log g
log δi < 1 + log
2qi+1 log(2qi+1 + 1)
2
Qi qi+1 log g
τ (ni ) − 2
+ 2 1 + log 2 . (10)
2qi+1 log(2qi+1 + 1)
log qi+1
Hence, log δi qi+1 , where the constant implied by the Vinogradov sym-
bol above depends only on g, implying that qi+1 must be bounded by
some effectively computable constant depending only on g. This concludes
the proof of Theorem 1.
4 Proof of Theorem 2
We assume that g is odd and 3 ≤ g ≤ 999, so that 3 ≤ q1 ≤ 31 by Lemma
8.
Claim 1: That νq1 (uq1 −1 ) ≤ 5 can be checked with Mathematica. In
particular, by Lemma 6, we have that if q1 g, then νq1 (φ(un )) ≤ 5.
Claim 2: τ (n/q1 ) ≤ νq1 (φ(un )) ≤ 6, and s ≤ 3.
Suppose first that q1 | g. Then, by Lemma 6,
log g log 1000
τ (n/q1 ) ≤ νq1 (φ(un )) ≤ νq1 (gun−1 ) = νq1 (g) ≤ ≤ = 6.
log q1 log 3
In the above, in fact νq1 (g) < 6 unless (q1 , g) = (3, 729). Then, for any q1 ,
by Claim 1, either q1 = 3 and τ (n/q1 ) ≤ 6, or τ (n/q1 ) ≤ 5. In particular,
τ (n) ≤ 2τ (n/q1 ) ≤ 12, which shows that s ≤ 3.
10

11. Claim 3: s ≥ 2.
Let us see indeed that for our particular case we cannot have s = 1. If
α
this were so, then n = q1 1 . Then each prime factor pj of un is primitive
for some divisor d > 1 of n, which is a power of q1 (again, this is because
gcd(un , g − 1) = 1). Thus, pj ≡ 1 (mod q1 ) for all j = 1, . . . , K, showing
that νq1 (φ(un )) ≥ K ≥ 14 (see Lemma 2 iii)), which contradicts the fact
that νq1 (φ(un )) ≤ 6. Hence, s ≥ 2.
Claim 4: α1 = 1 except when (α1 , q1 , g) = (2, 3, 729).
α
Put again, as in the proof of Theorem 1, Q1 = q1 1 . By Lemma 6 and
the fact that s ≥ 2, we have
α1 (α1 + 1) α
α1 (α1 + 1) ≤ τ (n/q1 1 ) ≤ νq1 (φ(un )).
2
By Claims 1 and 2 above, we know that νq1 (φ(un )) ≤ 5, except when
(α1 , q1 , g) = (2, 3, 729). So, α1 = 1 except for this case.
Note that, at any rate, since s ≥ 2, it follows that 2 ≤ τ (n/q1 ) ≤
νq1 (guq1 −1 ). A computation with Mathematica revealed 431 possibilities for
the pairs (q1 , g) in our range satisfying νq1 (guq1 −1 ) ≥ 2.
Claim 5: q2 ≤ 19.
The smallest left hand side in (9) computed over all the 432 possible
pairs (Q1 , g) has δ1 > 1.49 (it was obtained for g = 809, Q1 = q1 = 3 and
a = 2, for which the obtained value is > 1.495). Of course, we did not
factor all the numbers of the form (g Q1 − 1)/(g − 1). If q1 = 31, then the
smallest prime p1 ≡ 1 (mod q1 ) is 311. The number K of prime factors of
u31 satisfies therefore
log uq1 3 · 31 · log 10
K< < < 38;
log p1 log 311
hence,
37
φ(uq1 ) 1
a ≥2 1− > 1.7.
uq1 311
Similarly, using the fact that when q1 = 29 and 23 the first two primes
congruent to 1 (mod q1 ) are 59 and 233, and 47 and 139 respectively, and
3 · 29 · log 10 3 · 23 · log 10
< 37 and < 33,
log 233 log 139
11

12. we have that
36 32
φ(uq1 ) 1 1 1 1
a ≥ 2 min 1− 1− , 1− 1−
uq1 59 233 47 139
> 1.55,
whenever q1 ∈ {23, 29}. Thus, we have factored only the numbers uQ1 with
Q1 ≤ 19. We now use inequality (10) for i = 1 to obtain
ω(n1 ) Q1 q2 log g
log(1.49) < 1 + log
2q2 log(2q2 + 1)
τ (n1 ) − 2 2
Q1 q2 log g
+ 2 1 + log 2 .
2q2 log(2q2 + 1)
2
If q1 > 3, then Q1 = q1 ≤ 31. If q1 = 3, then Q1 = q1 = 9. Thus, Q1 ≤ 31
in both cases. We also saw in Claims 1 and 2 that τ (n1 ) ≤ τ (n/q1 ) ≤ 6, so
also ω(n1 ) ≤ 2. Hence,
1 31q2 log 999 2 2
31q2 log 999
log(1.49) < 1 + log + 2 1 + log 2 ,
q2 log(2q2 + 1) q2 log(2q2 + 1)
and this inequality does not hold when q2 ≥ 23.
4.1 The conclusion of the proof of Theorem 2
Thus, 3 ≤ q1 < q2 ≤ 19. The argument showing that α1 = 2 except if
(q1 , g) = (3, 729) now shows that α2 = 1. We are now able to show that
s = 2. Indeed, if it were not so, then we would have both τ (n/q1 ) ≥ 4
and τ (n/q2 ) ≥ 4. A quick computation with Mathematica shows that while
there are pairs (q, g) such that νq (guq−1 ) ≥ 4 in our ranges, there is no odd
g in [3, 999] that has the above property with respect to two different primes
3 ≤ q1 < q2 ≤ 19. Thus, either n = q1 q2 , or n = 9q2 and g = 729. To test
these last pairs, we proceeded as follows. First we have detected all pairs
(n, g) with n = q1 q2 with 3 ≤ q1 < q2 ≤ 19 and odd g ∈ [3, 999] such that
νqi (gun−1 ) ≥ 2 holds for both i = 1, 2. There are 2043 such pairs. For each
one of these we checked that ν2 (un−1 ) < 14. Similarly, when Q1 = 9 and
g = 729, the only possibility for q2 in our range such that νq2 (uq2 −1 ) ≥ 2 is
q2 = 11, but in this case n = 99 and ν2 (un−1 ) = 1 < 14. This finishes the
proof of Theorem 2.
12

13. References
[1] A. S. Bang, “Taltheoretiske Undersølgelser”, Tidskrift f. Math. 5
(1886), 70–80 and 130–137.
[2] G. L. Cohen and P. Hagis, “On the number of prime factors of n if
φ(n) | n − 1”, Nieuw Arch. Wisk. (3) 28 (1980), 177–185.
[3] R. K. Guy, Unsolved Problems in Number Theory, Springer, 2004.
[4] D. H. Lehmer, “On Euler’s totient function”, Bull. Amer. Math. Soc.
38 (1932), 745–751.
[5] F. Luca, “Fibonacci numbers with the Lehmer property”, Bull. Polish
Acad. Sci. Math. 55 (2007), 7–15.
[6] C. Pomerance, “On composite n for which φ(n) | n − 1, II”, Pacific J.
Math. 69 (1977), 177–186.
13